00:01
In this exercise, we have two point charges, a and b, where a is free to move on the x direction, and b is fixed to displace.
00:11
Besides, a and b are connected to each other by a massless and non -conducting wire here.
00:19
And also, they are subjected to an external magnetic field here in green.
00:26
I also wrote on the right -hand side some important quantities we must.
00:31
Keep in mind to solve the exercise.
00:34
So first, i wrote the charge of particle a, which is negative.
00:39
I wrote the charge of particle b, which is positive, and i wrote the external electric field e.
00:51
And here in parentheses, i'm emphasizing that it's pointing to the positive x direction.
00:56
And finally, i wrote the pulum constant, which will need to calculate the electric field generated by this point charges.
01:10
Okay, so question a asks us to calculate the tension on the wire.
01:17
So notice that since particle b is fixed, the tension on the wire should be equal to the force on particle a.
01:29
But the force on particle a is the force that the external field exerts, on particle a plus the force that the particle b exerts on particle a.
01:53
So in starting with the force generated by the external magnetic field, we have that the electric force is particle a's charge, which i'll write as just a.
02:15
Times the external magnetic field.
02:22
But since particle a is negative, has negative charge, we have that the force of the external field is pointing to the negative x direction here.
02:44
And it is, and it is, the value of charge a, which is, minus 6 .5 times 10 to the minus 6 quorum times the value of the electric field, 1 .8 5 times 10 to the 8 newton per quillum on the x direction.
03:16
If we substitute all these values, we find that the force that the external electric field exerts on the on particle a is just 1 .20 times 10 to the third newton.
03:42
So i'll highlight this value here because we'll need it later on.
03:50
Great, now we only have to know what is the force that particle b exerts on a.
04:00
Well, just like the other case, the value is the charge, of the particle, which is a, times the electric field that the particle b generates.
04:17
But to calculate what is the electric field that particle b generates, we must recall culum's law.
04:28
And from coulomb's law, we have that the electric field generated by charge b is k, which is the culum constant, times the chart of the particle b, which i'll write it just as b, over the distance between those two particles squared.
04:53
I forgot to mention that a distance between those two particles is equal to 2 .5 times 10 to the minus 2 meters, which is this distance here, the length of the wire.
05:16
Okay, now if we just substitute all these values inside this little equation here, we have that the force that the particle b exerts on particle a is equal to 8 .18 times 10 to 2 newton.
05:44
Notice that this force is pointing to the positive x direction...